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If $0 \to A \to B \to C \to 0$ is a short exact sequence of vector bundles on $X$, is it necessarily true that $[{\rm Sym}^k B] = \sum_{i=0}^k [{\rm Sym}^i A \otimes {\rm Sym}^{k-i} C]$ in the Grothendieck group of vector bundles on $X$? (Here ${\rm Sym}^k$ stands for the $k$-th symmetric power of a vector bundle).

Is it true that ${\rm Sym}^k$ induces a map from $K_0(X)$ to $K_0(X)$?

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    $\begingroup$ Yes. It is probably due to Dold, but reviewed in Section 3 of [Barwick–Glasman–Mathew–Nikolaus, K-theory and polynomial functors]. $\endgroup$
    – Z. M
    Commented Sep 20 at 16:44
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    $\begingroup$ The first statement follows from Hartshorne, Exercise II.5.16(c), which constructs a filtration on $\operatorname{Sym}^k B$ whose associated graded is $\bigoplus_{i=0}^k \operatorname{Sym}^i A \otimes \operatorname{Sym}^{k-i} C$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Sep 20 at 16:57
  • $\begingroup$ @R.vanDobbendeBruyn Thank you very much! Is the filtration in the solution to that exercise given by the images of ${\rm Sym}^i A \otimes {\rm Sym}^{k-i} B$ in ${\rm Sym}^k B$? $\endgroup$
    – Yellow Pig
    Commented Sep 20 at 17:41
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    $\begingroup$ It is much earlier than Hartshorne (of course). It should also be in Borel-Serre, for instance. $\endgroup$
    – Jason Starr
    Commented Sep 20 at 17:59

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